Board Previous Year Questions

How to Score Full Marks in the Board Exam

A complete bank of board-style questions with model answers for Chemical Kinetics. Write the formula first ($k=\dfrac{2.303}{t}\log\dfrac{[R]_0}{[R]}$, $t_{1/2}=\dfrac{0.693}{k}$, $k=Ae^{-E_a/RT}$), substitute with units, and box the answer. In first-order and Arrhenius numericals, examiners give method marks for the correct set-up.

1-Mark Questions (Definitions & Direct)

Q1. Define the order of a reaction. Answer: The order of a reaction is the sum of the powers of the concentration terms in the experimentally determined rate law. It can be zero, fractional or negative.

Q2. Define molecularity of a reaction. Answer: Molecularity is the number of reacting species that collide simultaneously in an elementary reaction. It is always a positive integer (1, 2 or 3).

Q3. Write the unit of the rate constant for a zero-order reaction. Answer: mol L$^{-1}$ s$^{-1}$ (the same as the unit of rate).

Q4. What is the half-life of a first-order reaction in terms of its rate constant? Answer: $t_{1/2} = \dfrac{0.693}{k}$; it is independent of the initial concentration.

1-Mark Questions (continued)

Q5. Write the Arrhenius equation and name its terms. Answer: $k = A e^{-E_a/RT}$, where $A$ is the frequency (Arrhenius) factor, $E_a$ the activation energy, $R$ the gas constant and $T$ the absolute temperature.

Q6. What is a pseudo-first-order reaction? Give one example. Answer: A reaction that is truly of higher order but appears first order because one reactant is present in large excess (its concentration stays nearly constant). Example: acid hydrolysis of ethyl acetate (water in excess).

Q7. How does a catalyst affect the activation energy of a reaction? Answer: A catalyst provides an alternative pathway of lower activation energy, thereby increasing the rate; it does not change $\Delta H$ or the equilibrium constant.

Q8. What is the effect of temperature on the rate constant? Answer: The rate constant increases with temperature, as described by the Arrhenius equation.

2-Mark Reasoning Questions

Q9. Distinguish between order and molecularity of a reaction (any two points). Answer: (i) Order is an experimental quantity and can be zero, fractional or negative; molecularity is theoretical and is always a positive integer. (ii) Order applies to both elementary and complex (overall) reactions; molecularity applies only to elementary reactions.

Q10. Why can the rate law not be written directly from the balanced chemical equation? Answer: The rate law depends on the reaction mechanism, particularly the slowest (rate-determining) step, not on the stoichiometric coefficients of the overall equation. The powers in the rate law are determined experimentally and may differ from the coefficients (they can even be fractional).

2-Mark Reasoning Questions (continued)

Q11. Why does the rate of a reaction increase sharply with a small rise in temperature, even though the average molecular speed increases only slightly? Answer: The rate depends not on the average speed but on the fraction of molecules with energy equal to or greater than the activation energy, given approximately by the Arrhenius factor $e^{-E_a/RT}$. This fraction is very sensitive to temperature, so a small rise in $T$ greatly increases the number of effective collisions and hence the rate.

Q12. Explain how a catalyst increases the rate of a reaction using the Maxwell-Boltzmann/Arrhenius idea. Answer: A catalyst lowers the activation energy $E_a$ by providing an alternative path. Since the fraction of molecules able to cross the barrier is $e^{-E_a/RT}$, a smaller $E_a$ means a larger fraction can react, so $k$ (and the rate) increases — without changing the temperature or $\Delta H$.

Q13. State the two conditions necessary for an effective collision in collision theory. Answer: (i) The colliding molecules must possess energy equal to or greater than the activation energy ($E_a$); and (ii) they must collide with the proper orientation. Only collisions satisfying both lead to product.

3-Mark Numericals

Q14. A first-order reaction has a rate constant of $1.15 \times 10^{-3}$ s$^{-1}$. How long will it take for the reactant to fall to one-fourth of its initial value? Answer: $t = \dfrac{2.303}{k}\log\dfrac{[R]_0}{[R]} = \dfrac{2.303}{1.15\times10^{-3}}\log 4 = 2002.6 \times 0.602 = \mathbf{1205\ s}$.

Q15. A first-order reaction takes 40 minutes for 30% decomposition. Calculate its rate constant. Answer: 30% decomposed means 70% remains, so $\dfrac{[R]_0}{[R]} = \dfrac{100}{70} = 1.4286$. $k = \dfrac{2.303}{40}\log 1.4286 = \dfrac{2.303}{40}\times0.1549 = \mathbf{8.92 \times 10^{-3}\ min^{-1}}$.

Q16. A first-order reaction is 50% complete in 20 minutes. Calculate its rate constant and half-life. Answer: $k = \dfrac{2.303}{20}\log 2 = \dfrac{2.303}{20}\times0.301 = 0.0347$ min$^{-1}$. $t_{1/2} = \dfrac{0.693}{k} = \dfrac{0.693}{0.0347} = \mathbf{20\ min}$ (consistent, since 50% complete = one half-life).

3-Mark Numericals (continued)

Q17. For a zero-order reaction, the concentration falls from 0.10 M to 0.075 M in 50 s. Calculate the rate constant and the half-life if $[R]_0 = 0.10$ M. Answer: $k = \dfrac{[R]_0 - [R]}{t} = \dfrac{0.10 - 0.075}{50} = 5 \times 10^{-4}$ mol L$^{-1}$ s$^{-1}$. $t_{1/2} = \dfrac{[R]_0}{2k} = \dfrac{0.10}{2 \times 5\times10^{-4}} = \mathbf{100\ s}$.

Q18. The rate of a reaction doubles when the temperature is raised from 298 K to 308 K. Calculate the activation energy. ($R = 8.314$ J K$^{-1}$ mol$^{-1}$) Answer: $\log\dfrac{k_2}{k_1} = \dfrac{E_a}{2.303R}\cdot\dfrac{T_2-T_1}{T_1T_2}$, so $\log 2 = \dfrac{E_a}{2.303\times8.314}\cdot\dfrac{10}{298\times308}$. Solving, $E_a = \mathbf{52.9\ kJ\ mol^{-1}}$.

Q19. The half-life of a radioactive isotope ($^{14}$C) is 5730 years. An archaeological sample contains 25% of the original $^{14}$C. Find its age. Answer: 25% remaining $= (1/2)^2$, so 2 half-lives have elapsed. Age $= 2 \times 5730 = \mathbf{11460\ years}$.

3-Mark Numericals (continued)

Q20. A first-order reaction has $k = 0.0693$ min$^{-1}$ and starts at $[R]_0 = 1.0$ M. Find $[R]$ after 10 minutes. Answer: $\log\dfrac{[R]_0}{[R]} = \dfrac{kt}{2.303} = \dfrac{0.0693\times10}{2.303} = 0.301$. So $\dfrac{[R]_0}{[R]} = 2.0$, giving $[R] = \mathbf{0.50\ M}$.

Q21. An Arrhenius plot of $\ln k$ versus $1/T$ has a slope of $-6000$ K. Calculate the activation energy. ($R = 8.314$) Answer: slope $= -E_a/R$, so $E_a = 6000 \times 8.314 = 49884$ J $= \mathbf{49.9\ kJ\ mol^{-1}}$.

Q22. Show that for a first-order reaction the time required for 99% completion is twice that required for 90% completion. Answer: $t_{99\%} = \dfrac{2.303}{k}\log\dfrac{100}{1} = \dfrac{2.303}{k}(2)$; $t_{90\%} = \dfrac{2.303}{k}\log\dfrac{100}{10} = \dfrac{2.303}{k}(1)$. Therefore $t_{99\%} = 2\,t_{90\%}$.

3-Mark Numericals (continued)

Q23. For a first-order reaction, $[R]$ falls from 0.80 M to 0.20 M in 60 s. Calculate the rate constant. Answer: $k = \dfrac{2.303}{t}\log\dfrac{[R]_0}{[R]} = \dfrac{2.303}{60}\log\dfrac{0.80}{0.20} = \dfrac{2.303}{60}\log 4 = \dfrac{2.303}{60}\times0.602 = \mathbf{0.0231\ s^{-1}}$.

Q24. Calculate the half-life of a first-order reaction whose rate constant is $1.386 \times 10^{-3}$ s$^{-1}$. Answer: $t_{1/2} = \dfrac{0.693}{k} = \dfrac{0.693}{1.386\times10^{-3}} = \mathbf{500\ s}$.

Q25. For the rate law Rate $= k[A]^2[B]$, what happens to the rate if the concentration of A is doubled and that of B is halved? Answer: New rate $\propto (2)^2 \times (1/2)^1 = 4 \times 0.5 = 2$. The rate doubles.

3-Mark Numericals & Reasoning

Q26. A zero-order reaction has $k = 0.02$ mol L$^{-1}$ s$^{-1}$ and $[R]_0 = 0.50$ M. Find $[R]$ after 10 s and its half-life. Answer: $[R] = [R]_0 - kt = 0.50 - (0.02)(10) = 0.30$ M. $t_{1/2} = \dfrac{[R]_0}{2k} = \dfrac{0.50}{0.04} = \mathbf{12.5\ s}$.

Q27. The rate constant of a reaction is $3.0 \times 10^{-4}$ s$^{-1}$. State its order and calculate its half-life. Answer: The unit s$^{-1}$ indicates a first-order reaction. $t_{1/2} = \dfrac{0.693}{3.0\times10^{-4}} = \mathbf{2310\ s}$.

Q28. Define rate of reaction and write the rate expression for $2\text{N}_2\text{O}_5 \rightarrow 4\text{NO}_2 + \text{O}_2$. Answer: Rate of reaction is the change in concentration of a reactant or product per unit time. Rate $= -\dfrac{1}{2}\dfrac{d[\text{N}_2\text{O}_5]}{dt} = +\dfrac{1}{4}\dfrac{d[\text{NO}_2]}{dt} = +\dfrac{d[\text{O}_2]}{dt}$.

3-Mark Numericals (continued)

Q29. The rate constant doubles when temperature rises from 300 K to 310 K. Calculate the activation energy. ($R = 8.314$) Answer: $\log 2 = \dfrac{E_a}{2.303\times8.314}\cdot\dfrac{10}{300\times310}$, giving $E_a = \mathbf{53.6\ kJ\ mol^{-1}}$.

Q30. A first-order reaction is 75% complete in 60 minutes. Find its half-life. Answer: 75% complete means 25% remains $= (1/2)^2$, so 2 half-lives elapse in 60 min. Therefore $t_{1/2} = 60/2 = \mathbf{30\ min}$.

Q31. Sr-90 has a half-life of 28.1 years. What fraction of a sample remains after 84.3 years? Answer: Number of half-lives $= 84.3/28.1 = 3$. Fraction remaining $= (1/2)^3 = \mathbf{1/8}$ (12.5%).

5-Mark / Long-Answer Questions

Q32. Derive the integrated rate equation for a first-order reaction. Answer: For a first-order reaction $R \rightarrow P$, the rate is $-\dfrac{d[R]}{dt} = k[R]$. Rearranging, $\dfrac{d[R]}{[R]} = -k\,dt$. Integrating between $[R]_0$ (at $t=0$) and $[R]$ (at time $t$): $\ln[R] - \ln[R]_0 = -kt$, i.e. $\ln\dfrac{[R]_0}{[R]} = kt$. Converting to base-10 logarithm: $k = \dfrac{2.303}{t}\log\dfrac{[R]_0}{[R]}$. A plot of $\log[R]$ versus $t$ is a straight line of slope $-k/2.303$.

Q33. Explain the Arrhenius equation and how it is used to find the activation energy graphically. Answer: The Arrhenius equation $k = A e^{-E_a/RT}$ relates the rate constant to temperature. Taking logarithms: $\ln k = \ln A - \dfrac{E_a}{RT}$. A plot of $\ln k$ against $1/T$ is a straight line with slope $-\dfrac{E_a}{R}$ and intercept $\ln A$. Thus the activation energy is obtained from $E_a = - (\text{slope}) \times R$, and the frequency factor from the intercept.

5-Mark / Long-Answer Questions (continued)

Q34. Derive the expression for the half-life of a zero-order reaction and state how it depends on the initial concentration. Answer: For a zero-order reaction, $[R] = [R]_0 - kt$. At the half-life, $[R] = [R]_0/2$. Substituting: $\dfrac{[R]_0}{2} = [R]_0 - k\,t_{1/2}$, so $k\,t_{1/2} = \dfrac{[R]_0}{2}$, giving $t_{1/2} = \dfrac{[R]_0}{2k}$. Hence the half-life of a zero-order reaction is directly proportional to the initial concentration (unlike a first-order reaction, whose half-life is independent of $[R]_0$).

Q35. The decomposition of N$_2$O$_5$ follows first-order kinetics with $k = 2.303 \times 10^{-3}$ s$^{-1}$. (a) What fraction remains after 1000 s? (b) What is its half-life? Answer: (a) $\log\dfrac{[R]_0}{[R]} = \dfrac{kt}{2.303} = \dfrac{2.303\times10^{-3}\times1000}{2.303} = 1$, so $\dfrac{[R]_0}{[R]} = 10$, i.e. 0.1 (10%) remains. (b) $t_{1/2} = \dfrac{0.693}{2.303\times10^{-3}} = \mathbf{301\ s}$.

Q36. Explain collision theory and the role of the steric (probability) factor. Answer: According to collision theory, molecules must collide to react, but only collisions with energy $\ge E_a$ and the correct orientation are effective. A more complete expression is $\text{Rate} = Z_{AB}\,P\,e^{-E_a/RT}$, where $Z_{AB}$ is the collision frequency and $P$ is the steric (probability) factor accounting for the fraction of collisions with proper orientation. This helps explain why the observed rate may be lower than the rate predicted from collision frequency alone.